Answer:
In chemistry, neutralization or neutralisation is a chemical reaction in which acid and a base react quantitatively with each other. In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in the solution.Neutralization reactions are the reaction between acid and base. The products formed are water and salt. It is called so because the acid and base neutralize each other to form water and salt.Hint: The neutralization reaction is the one in which an acid reacts with an equimolar amount of base to give salt and water. The example could be a reaction between any strong acid and a base. The sodium chloride formed is a result of neutralization reaction.
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The oxidation state of the elements in the compounds are:
CoH₂:
FeBr₃:
<h3>What is the oxidation states of the elements in the given compounds?</h3>
The oxidation states of the elements in each of the given compounds is determined as follows:
Cobalt dihydride, CoH₂
Co = +2
H = -1
Iron (iii) bromide, FeBr₃
Fe = +3
Br = -1
In conclusion, the oxidation state of the elements are charges they have in the compound.
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We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
The answer is letter A definitively .
Answer: You started with 8
Explanation: the amount of products is equal to the amount of reactants
